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Commercially available concentrated hydrochloric acid contains 38% HCl by mass. (a) What is the molarity of this solution ? The density is 1.19 g cm^(-3) (b) What volume of concentrated hydrochloric acid is required to make 1.00 L of 0.10 M HCl ? |
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Answer» SOLUTION :CALCULATION of molarity. `38%` HCL by mass means that 38 g of HCl are present in 100 g of the solution. `"Volume of 100 g of the solution"=("Mass ")/("Density")=(100g)/(1.19"g cm"^(-3))=84.03cm^(3)=0.0840L` Molar mass of HCl = `36.5"g mol"^(-1)` `therefore"38 g HCl"=(38)/(36.5)"moles"=1.04" moles"` `"Molarity of solution "=(1.04"moles")/(0.0840L)=12.38"mol L"^(-2)`, i.e., Molarity of the solution = 12.38 M (b) Calculation of volume of conc. HCl for 1.00 L of 0.10 M HCl. Applying molarity equaton, we have `{:(M_(1)xxV_(1),=,""M_(2)xxV_(2)),("(conc. HCl)",,"(1.0 L of 0.10 M HCl)"),(12.38xxV_(1),=,""0.10xx1.0):}` `"or"V_(1)=(0.1)/(12.38)L=(0.1)/(12.38)xx1000cm^(3)=8.1cm^(3)` |
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